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97 Cards in this Set
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Equilibrium
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Rate fwd = Rate bckwd Both of these reactions are happening in the solution simultaneously (at the same time). The Rate of precipitation = The Rate of dissolving |
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Dynamic Equilibrium |
Macroscopically: Nothing appears to be happening Microscopically: Both reactions are occurring furiously at the same time |
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Reactions proceed to Equilibrium |
ex: Table salt can be fully dissolved (rxn runs to completion) Or the dissolution will stop when the solution is saturated |
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Law of Mass Action |
For the rxn: aA(aq) + bB(s) ⇌ cC(g) + dD(l) K = ((aC)^c(aD)^d)/((aA)^a(aB)^b) This is called the Law of Mass Action Where a is called the activity And K is called the Equilibrium Constant So K = (aProducts)/(aReactants) to the power of the rxn stoichiometry |
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Activity (a) |
activity = (concentration)/(reference) For the rxn: aA(aq) + bB(s) ⇌ cC(g) + dD(l)\ activity represents how active the species is in the chemical equation once the reaction has reached equilibrium |
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Equilibrium Constant (K) |
For the rxn: aA(aq) + bB(s) ⇌ cC(g) + dD(l) K = ((aC)^c(aD)^d)/((aA)^a(aB)^b) K = ([C]^c (1)^d) /([A]^a(1)^b) (We don’t include the 1 in the expression for K) K= ([C]^c/[A]^a) K is a ratio of activities based on reaction stoichiometry The value of K has NO units For a given reaction at constant temp, K is CONSTANT |
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Equilibrium Constant (K)(cont.) |
For the rxn: aA(aq) + bB(s) ⇌ cC(g) + dD(l) K= ([C]^c/[A]^a) If you start with only A and B, the concentrations will change until the equilibrium condition is met and K is achieved. If you start with only C and D, the concentrations will change until the same equilibrium condition is met and K is achieved. It doesn’t matter if you start with only reactants, or only products, the reaction will proceed to the same point! k has no units btw in case u were wondering |
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When K >> 1 |
Product favored reaction. Rxn proceeds forward At equilibrium there is a lot of product and very little reactant. CH3O2 + NO2 ⇌ CH3O2NO2 K = 1.2 x 10^33 |
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When K << 1 |
Reactant favored reaction. Rxn proceeds backward At equilibrium there is a lot of reactant and very little product O3 + NO ⇌ NO2 + O2 K = 5.8 x 10^‐34 |
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What does K mean? |
K is the measure of how far a reaction will proceed in the direction it has been written. K = [Products]/[Reactants] By convention: if K is on the order of 10^8 or higher we say the reaction goes to completion |
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define large & small k |
Large K means that at EQUILIBRIUM you have lots of product and not much reactant Small K means that at EQUILIBRIUM the concentration of products is smaller than the concentration of reactants. |
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K depends on the chemical equation |
The numerical value of K depends on how the reaction has been written The numerical value of K for one reaction can be related to the numerical value of K for another reaction without even knowing the species concentrations. |
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A very large value of K (10^8): |
Rxn runs to completion [products] > [reactants] The equilibrium lies toward the product side (*A rxn that effectively runs to completion is really NOT over. There is still some product being converted into reactant, but it is a very small amount) |
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A very small value of K (10^‐12): |
Rxn effectively doesn’t happen [reactants] > [products] The equilibrium lies toward the reactant side (*A rxn that eectively doesn’t happen in fact does happen, but to a very small extent. There is some reactant being converted into product, but it is virtually undetectable.) |
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Rule 1: Reversing reactions |
Kforward = [CH3O2NO2]/[CH3O2][NO2] Kback = [CH3O2][NO2]/[CH3O2NO2] Kf = 1/Kb When you reverse a rxn the value of Krev is the inverse value of K for the original rxn. |
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Rule 2: Stoichiometric coecients |
I2(g) + H2(g) ⇌ 2HI(g) K1 = 0.5 K1 = [HI]^2/[I2][H2] X ½ 0.5I2(g) + 0.5H2(g) ⇌ HI(g) K2 = 0.71 K2 = [HI] /[I2]^1/2[H2]^1/2 K2 = (K1)½= (0.5) ½ = 0.71 2I2(g) + 2H2 (g) ⇌ 4HI(g) K3 = 0.25 K3 = [HI]⁴ /[I2]²[H2]² K3 = (K1)²= (0.5)² = 0.25 When you multiply a reaction by a factor N, K is raised to the power N → Knew = Kⁿ |
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Rule 3: Adding reactions |
Knew = K1K2 = [HI]²/[H2][I2]× [CO][I2]/[I2CO] = [HI]²[CO][I2]/[I2CO][H2][I2] = [HI]²[CO] /[I2CO][H2] When you add 2 rxns together the new K is the product of the K’s for the rxns you are adding. |
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Reaction Quotient (Q) |
Non-‐‐equilibrium REACTION QUOTIENT (Q): Solution concentration at any arbitrary set of conditions Q = ([HI]2non-eq)/([H2]non-eq[I2]non-eq) Q: concentrations at any point in the reaction → many possible values |
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Predicting the direction of a reaction |
compare the values of Q and K to predict the direction of a reaction If Q < K : Reaction proceeds forward (RIGHT) If Q > K: Reaction proceeds backwards (LEFT) If Q = K: Reaction is at equilibrium |
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IF the System has too much reactant Q ___ K what does it do to reach equi? |
is less than shifts to right (As the reaction proceeds right [P] increases and [R] decreases) |
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if System is at equilibrium, then Q ___ K |
equals |
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IF system has too much product then Q ___ K which way does it shift to reach equi? |
is greater than shifts left |
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The direction of a reaction depends on two things: |
1. The starting conditions (non-equilibrium concentrations) → value of Q 2. The concentrations at equilibrium → value of K |
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________ to the power of the rxn stoichiometry AT ANY OTHER POINT |
Q = [Products]/[Reactants] |
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_________to the power of the rxn stoichiometry AT EQUILIBRIUM |
K = [Products]/[Reactants] |
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Le Chatelier’s principle |
Reactant particles ⇌ Product particles When a system at equilibrium is changed (stressed) the system responds by shifting back to equilibrium Equilibrium (Q = K) STRESS → Non-‐‐Equilibrium (Q ≠ K) Response → Equilibrium (Q = K) Le Chatelier’s principle allows us to predict this response The system will always “relieve the stress” by proceeding either forward or backward until equilibrium is re-established |
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Equilibrium position |
Once equilibrium has been re-established the position of the equilibrium (equilibrium position) has changed even though the ratio of concentrations is once again the same. I2(l) + H2(g) ⇌ 2HI(g) K = 0.5 Equilibrium position = the set of concentrations for Q or K: { [H2], [HI] } Q = [HI]2/[H2] = (0.71)^2/(1) = 0.5 At equilibrium: [H2] = 1.0 M and [HI] = 0.71 M Equilibrium position = { 1.0, 0.71 } Q=K |
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Le Chatelier’s principle |
When a system at equilibrium is changed (stressed) the system responds by shifting back to equilibrium. The system will always “relieve the stress” by proceeding either forward or backward until equilibrium is re-established |
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le chat There are 4 Types of stress that can be applied: |
1. Change in concentration 2. Change in temperature 3. Change in volume/pressure (gaseous species only) 4. Addition of an inert gas or adding a catalyst |
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Increasing concentration (chat) N2(g) + 3H2(g) ⇌ 2NH3(g) K = 0.44 At equilibrium: [NH3] = 0.71 M If we add reactant (N2 or H2) to the system |
There is now more reactant than there was at equilibrium Q < K The equilibrium will shift to the right, towards products, to consume the extra reactant Shift right until Q = K |
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Increasing concentration (chat) N2(g) + 3H2(g) ⇌ 2NH3(g) K = 0.44 At equilibrium: [NH3] = 0.71 M If we add product (NH3) to the system |
There is now more product than there was at equilibrium Q > K The equilibrium will shift to the left, towards reactants, to consume the extra product Shift left until Q = K |
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Decreasing concentration (Le Chatelier’s Principle allows us to predict these changes without ever doing any calculations for Q) If we remove reactant from the system |
There is now less reactant than there was at equilibrium The equilibrium will shift to the left, towards reactants, to make more of the reactant SHIFT LEFT |
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If we remove product from the system |
There is now less product than there was at equilibrium The equilibrium will shift to the right, towards products, to make more of the product SHIFT RIGHT |
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If the concentration of reactants or products is changed, |
the reaction will proceed in the direction that will compensate for the change |
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Addition of matter: (k remains the same) |
Shift away |
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Removal of matter: (k remains the same) |
Shift toward |
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Change in temperature - exothermic (1) What happens if we increase the temperature of the reaction??? (2) What happens if we decrease the temperature of the reaction??? |
C (graphite) + O2(g) ⇌ CO2(g) + heat ΔH < 0 This reaction releases heat - EXOTHERMIC! Heat can be thought of as a product in an exothermic rxn. (1) The Equilibrium will shift to consume the added heat! Equilibrium shifts toward the reactants (2) The Equilibrium will shift to produce more heat! Equilibrium shifts toward the product |
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Change in temperature -‐‐ endothermic (1) What happens if we increase the temperature of the reaction??? (2) What happens if we decrease the temperature of the reaction??? |
heat + CaCO3 (s) ⇌ CaO (s) + CO2 (g) ΔH > 0 This reaction absorbs heat -‐‐ ENDOTHERMIC Heat can be thought of as a reactant in an endothermic rxn. (1) The Equilibrium will shift to consume the added heat! Equilibrium shifts toward the products (2) The Equilibrium will shift to produce more heat! Equilibrium shifts toward the reactant |
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ENDO: Heat is a __________ |
reactant |
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EXO: Heat is a __________ |
product |
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When T changes, ___________(what else changes) |
K changes (When temperature changes the equilibrium position shifts AND the numerical value of K is changed....K is a function of temperature: K(T)) |
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Consider the endothermic reaction: heat + C(s) + CO2(g) ⇌ 2CO(g) If such a system at equilibrium is heated, equilibrium will |
shift to the right; the value of K increases |
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K decreases when the temperature is raised (for what kind of reaction) |
for exothermic reactions: |
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The partial pressure (p) of a gas is PV = nRT |
the pressure exerted on the wall by only that gaseous species. pNO2 = (nNO2/V)RT |
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Daltons Law of Partial Pressures |
In a system with multiple gaseous species the total pressure exerted on the walls of the container is the sum of the pressures that each gas would exert if it were alone. (For this mixture of gases in a closed container: Ptotal = pN2 + pO2 + pNO2) |
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Increase volume, _________ pressure To compensate for this change, the equi shifts to the side with _______________ N2(g) + 2O2(g) ⇌ 2NO2(g) From 2 moles to 3 moles (________ pressure) When volume is increased this reaction shifts to the _______ (in this specific rxn) |
decrease more moles (to increase pressure) increasing LEFT |
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Decrease volume, ___________ pressure To compensate for this change, the equi shifts to the side with _______________ N2(g) + 2O2(g) ⇌ 2NO2(g) From 3 moles to 2 moles (________ pressure) When volume is decreased this reaction shifts to the _______ (in this specific rxn) |
increase less moles (to decrease pressure) decreasing RIGHT |
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Changes in volume only aect gaseous systems If we increase the volume: pressure __________ Shift is towards the side with _____ moles of gas |
decreases more |
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Changes in volume only aect gaseous systems If we decrease the volume: pressure __________ Shift is towards the side with _____ moles of gas |
increases less |
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When the moles of gas are the same for reactant and product there is ________ in equilibrium When volume changes the equilibrium position shifts.....& the numerical value of K (does what)________________ |
no shift remains unchanged |
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IF there's no moles of gas and volume increases/decreases then pressure ________ |
has no change (Since no moles of gas) |
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Consider the reaction C(s) + O2(g) ⇌ CO2(g) If I add C(s), the reaction will shift _____, and if I reduce the size of the reaction ask, the reaction will shift ____. |
neither left nor right, neither left nor right |
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Consider the reaction PF5(g) ⇌ PF3(g) + F2(g) If the size of the reaction vessel were to double, the reaction would_____ and K would_____. |
Shift right, remain unchanged |
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Addition of a noble gas (group 8) Adding the noble gas changes the ___________ of the system, but it does not interact with the _________ in the system The noble gas does not aect the _________ or _______ of the reactants or the products. |
overall pressure other gases partial pressures concentrations There will be NO SHIFT in equilibrium and no change in K |
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A catalyst speeds up the rate of a reaction Both the forward and backward reaction rates are changed simultaneously When you add a catalyst the equilibrium position ___________ and the numerical value of K ___________ |
remains unchanged remains unchanged |
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What does a catalyst actually do? |
It helps a reaction occur by lowering the activation energy. (For example: enzymes are catalysts) |
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Consider the exothermic reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g). 1. What will happen to the equilibrium if a catalyst is added? 2. Which combination of changes would always shift the equilibrium to the left? A. Increasing temperature and increasing volume B. Decreasing temperature and increasing volume C. Increasing temperature and decreasing volume D. Decreasing temperature and decreasing volume |
1. no shift in equi 2. C. Increasing temperature and decreasing volume |
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All solutions have acidic and/or basic character. 3 Theories that dene Acids and Bases: |
ACIDS 1. Arrhenius acids: contain H+ 2. Bronsted-Lowry acids: H+ donors 3. Lewis acids: electron pair acceptors BASES 1. Arrhenius bases: contain OH- 2. Bronsted-Lowry bases: H+ acceptors 3. Lewis bases: electron pair donors |
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1. Arrhenius Theory The Hydronium Ion |
H+ ions (protons) are not stable in aqueous solution The protons bind to water to form H3O+ (hydronium) ions. An acid always reacts with water to produce hydronium ions HA(aq) → H+(aq) + A¯ˉ(aq) |
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Bronsted-‐‐Lowry deniCon – Base Acid: Base: |
H+ Donor H+ Acceptor |
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1. Identify the acid, the base, the conjugate acid and the conjugate base. 2. Label the appropriate acid-base conjugate pairs for this reaction. HNO2 + H2O ⇌ H3O+ + NO2- |
CONJUGATE ACID-BASE PAIR ACID: HNO2 BASE: NO2- CONJ ACID-BASE PAIR ACID: H3O+ BASE: H2O |
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An acid will always donate a proton to the base it is reacting with to generate its conjugate base. A base will always accept a proton from the acid it is reacting with to generate its conjugate acid. |
:))) |
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An AMPHOTERIC molecule can act as |
either an acid or a base |
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Water is an amphoteric species, which means _____________________ and _______________________ |
Water is a base when it reacts w an acid Water is an acid when it reacts w a base |
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Auto-ionization of water |
An AMPHOTERIC molecule can act as either an acid or a base This means that one water molecule can ionize anotherwater molecule → Self-ionization of water |
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H2O + H2O ⇌ OH- + H3O+ Kw = _______________________________ |
[H3O+][OH-] = 1 x 10-14 at 298K |
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Acids react with water (and dissociate) |
We can use the equilibrium constant (K) to determine the relative strengths of acids and bases |
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The larger the value of Ka, the _______ the acid (____ dissociation) |
stronger more |
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HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) |
Ka = [H3O+ ][A-]/[HA] |
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Ka > 1 → [H3O+] > [HA] We call this type of acid |
Strong Acid |
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Ka < 1 → [H3O+] < [HA] We call this type of acid |
Weak Acid |
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Acid dissociation determines strength Strong acids _________________ in water Weak acids _________________ in water |
disassociate completely do not disassociate completely
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Ka is a measure of acid dissociation in water Strong acids: _____ Weak acids: _____ |
Ka > 1 Ka < 1 |
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Ka Strong Acid: Weak Acid: |
Ka > 1 0 < Ka < 1 |
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Ka values for all weak acids are tabulated |
The strongest weak acids dissociate the most in waterand thus have the largest Ka values |
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Strong versus weak bases Kb >>> 1 no BOH in solution We call this type of base ____________ Kb < 1 → [OH-] < [B] We call this type of base ___________ |
(Strong bases dissociate completely in water) BOH: Strong Base B: Weak Base |
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Strong Acids: |
Hydrochloric acid (HCl) Hydrobromic acid (HBr) Hydroiodic acid (HI) (check this one out) Nitric acid (HNO3) Chloric acid (HClO3) Perchloric acid (HClO4) Sulfuric acid (H2SO4) (only the rst proton) |
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Strong Bases: |
Sodium hydroxide (NaOH) Potassium hydroxide (KOH) Lithium hydroxide (LiOH) Strontium hydroxide (Sr(OH)2) Calcium hydroxide (Ca(OH)2) Barium Hydroxide (Ba(OH)2) Group 1 and a few group 2 metal hydroxides |
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Which of the following unknown species is the strongest base? A. Base 1 (Kb = 10-2) B. Base 2 (Kb = 10‐4) C. Base 3 (Kb = 10-6) D. Base 4 (Kb = 10-8) |
A. Base 1 (Kb = 10‐2) |
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How acidic or basic a solution is depends on ___________________________ |
the concentration of protons (H+) in the solution. |
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pH = -log[H3O+] |
If y = log10x, then x = 10^y |
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pH = -log[H3O+] pH = -log[H+] |
Remember that [H3O+] = [H+] |
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1) What is the pH of a solution with [H+] = 0.426 M? 2) What is the [H3O+] of a solution with pH = 6.34? |
1) .37 2) 4.57 x 10^-7 |
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[H+] depends on equilibrium position |
To calculate pH we need to know the concentration of the H3O+ ions in the solution at equilibrium |
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pH + pOH = In pure water (neutral pH): [H3O+] = [OH-] Kw = [H3O+][OH-] = |
14 10-14 at 298K (only true @ this temp) |
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What is the pH of 0.01 M solution of KOH? What is the hydronium ion concentration of a HOBr solution at pH 3.34? What is the hydroxide ion concentration of a HOBr solution at pH 3.34? |
12 4.6 x 10-‐‐4 M 2.2 x 10-‐‐11 M |
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Weak Acid pH Calculations |
0 < Ka < 1 pH = -log[H3O+] ≠ -log[HA] A weak acid does not dissociate completely [H3O+] must be determined using the ICE chart HA(aq) + H2O(l) ⇌ H3O+(aq) + A‐(aq) Ka = [H3O+][A-]/[HA] |
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Weak Base pOH Calculations |
0 < Kb < 1 pOH = -log[OH-] ≠ -log[B] A weak base does not react completely [OH-] must be determined using the ICE chart B(aq) + H2O(l) ⇌ BH+(aq) + OH-‐‐(aq) Kb = [OH‐][BH+]/[B] |
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Salts dissociate in water to form ions What is a SALT? |
Ionic compound: metal + non-metal When soluble salts are dissolved in water theydissociate into their cons-tuent ions (ca-ons and anions) spectator ions and pH active |
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Spectator ions |
Conjugate species to STRONG acids and bases Spectator ions are not pH active (spectator ions don’t react with water) |
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pH active species |
Conjugate species to WEAK acids and bases (will react with water) |
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(Predicting the acidity of salts) We can predict the acidity of a salt by dissociating it and then identifying the pH active species in solution Examples Neutral salt: (NaCl) Acidic salt: (NH4Cl) Basic salt: (NaNO2) |
2 spectator ions (conjugate acid to a strong base and conjugate base to a strong acid) Na+ <-- NaOH Cl- <-- HCl spectator is an anion pH active: CONJUGATE ACID NH4+ <-- NH3 spectator is a cation pH active: CONJUGATE BASE NO2 <‐‐ HNO2 |
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An aqueous solu1on of Mg(CN)2 An aqueous solu1on of NH4Cl is An aqueous solu1on of KNO3 is |
Basic Acidic Neutral |
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(Remember A/B conjugate pairs) For any A/B conjugate pair: |
Ka x Kb = Kw = 1 x 10-14 pKa + pKb = pKw = 14 |
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The smaller the value of pH, the _______ the acid. The smaller the value of pKa, the ______ the acid. The smaller the value of pOH, the _______ the base. The smaller the value of pKb, the _______ the base. |
stronger stronger stronger stronger |
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1. Consider a chemical reaction with K = 1.6 x 10^18. Which of the following statements is TRUE? 2. The value of K for the reaction A ⇌ B is 1.4 × 10-15. At equilibrium which of the following statements is true? |
1. At equilibrium, the concentration of the products would be much greater than the concentrations of the reactants. 2. The amount of A is much larger than the amount of B. |
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Consider the reaction: O2(g) + 2H2(g) ⇌ 2H2O(g) K1 = 1.4 x 10^11 and use it to determine the numerical value of K for the reactions below. 1. 2H2O(g) ⇌ O2(g) + 2H2(g) 2. ½O2(g) + H2(g) ⇌ H2O(g) 3. 2I2(g) + 2H2O(g) ⇌ 4HI(g) + O2(g) (To do this you also need to consider I2 + H2 ⇌ 2HI K2 = 0.5) |
1. Knew = 1/(K1) = 7.1 x 10^-12 2. Knew = (K1)1/2 = 3.7 x 10^5 3. Multiply K2 rxn by 2 and add to the inverse of rxn K1 Knew = (K2)2 * (1/K1) = 1.8 x 10^-12 |
